At $0.1 a t m$ and $25^{\circ} C$ , compressibility factor of four gases $A, B, C$ and $D$ are $0.25, 1.3, 1 a n d 0.7$ respectively. The gas which is most easily liquifiable at $25^{\circ} C,$ will be.

D (g)

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C (g)

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B (g)

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A (g)

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Solution

The correct option is D $A (g)$
Less will be the value of $Z$ , more will be negative deviation, hence more will be the attractive forces.
$Z < 1$ (negative deviation)
Attractive force will dominate so, $A (g)$ will be more compressible / liquifiable.

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Similar questions

0.1 a t m

25^{\circ} C

A, B, C

D

0.25, 1.3, 1 a n d 0.7

25^{\circ} C,

25^{\circ} C, 10^{\circ} C, - 80^{\circ} C a n d 15^{\circ} C

Gases	$A$	$B$	$C$	$D$
$T_{B}$	$120 K$	$25 K$	$500 K$	$410 K$

26^{\circ} C, 10^{\circ} C, - 80^{\circ} C

15^{\circ} C

O_{2}, N_{2}, N H_{3}

1.360, 1.390, 4.170

L^{2} a t m m o l^{- 2}

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